Question

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooling system holds 5.60 gal, what is the boiling point of the solution? For the calculation, assume that at normal filling conditions, the densities of engine coolant and water are 1.11 g/mL and 0.998 g/mL respectively. Also, assume that the engine coolant is pure ethylene glycol ( HOCH 2 CH 2 OH ) , which is non‑ionizing and non‑volatile, and that the pressure remains constant at 1.00 atm. The boiling‑point elevation constant for water will also be needed.

Answer 1

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ $\frac{5.60}{2}gallons = 2.80 gallons$

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = $(M__1}) = Density*Volume$

                          $= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}$

                         $= 11659.06grams$

On the other hand; the mass of water = $(M__2})= Density*Volume$

                         $= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}$

                        $= 10577.95 grams$

Molarity = $\frac{massof solute*1000}{molarmassof solute*massofwater}$

              =  $\frac{11659.06*1000}{62.07*10577.95}$

              = 17.757 m

              ≅ 17.76 m

∴  the boiling point of the solution is calculated using the  boiling‑point elevation constant for water and the Molarity.

$\Delta T_{boiling} = k_{boiling}M$

where,

$k_{boiling}$ = 0.512 °C/m

$\Delta T_{boiling}$ =  100°C + 17.56 × 0.512

              = 109.09 °C